Disclaimer, I'm a physics student, so some of my arguments might be a little bit sloppy. I'm trying to find a somewhat straightforward general argument to explain why we need central extensions when working with quantum symmetry groups. Here's what I think I've more or less understood (most of my current understanding is based on Schottenloher's book on CFT and this post):
Since quantum systems have an inherent $\mathrm{U}(1)$ phase symmetry, we can allow for projective representations of our symmetry group on the Hilbert space $\mathcal{H}$ on top of the "usual" ones. To do this we extend the group $G$ to the central extension $E$, given by
$$ 1 \longrightarrow \mathrm{U}(1) \longrightarrow E \longrightarrow G \longrightarrow 1 $$
which corresponds to the central extension of the Lie algebra
$$ 0 \longrightarrow \mathfrak{u}(1)\cong\mathbb{R} \longrightarrow \mathfrak{e} \longrightarrow \mathfrak{g} \longrightarrow 0. $$
This seems reasonably intuitive, since we are somehow associating a (more or less) arbitrary phase to each $g\in G$, corresponding to the additional phase we allow for in the group multiplication.
As far as I understand, if these extensions are deemed "trivial" meaning that $E$ always factorizes into $\mathrm{U}(1) \times G$ (or equivalently, if $\mathfrak{e}\cong\mathfrak{g}\oplus\mathbb{R}$), then the central extension is effectively redundant, since we are in some sense free to just choose $1 = e^{i\varphi} \in \mathrm{U}(1)$, which means we can always modify a projective representation to be a normal one (?).
But what happens when $G$ doesn't admit any nontrivial central extensions, but isn't simply connected? Of course the correct thing would be to consider the univeral cover $\tilde{G}$ of $G$, but how does this follow from the previous argument (if at all)? Do we require $G$ to be the universal cover to begin with, and if so, is there some intuitive way of understanding this step?
